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Stochastic stability when type size difference is large

4. Stochastic Stability with Double Mutations

4.2. Stochastic stability when type size difference is large

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no more move mutations than MRT in the stochastic potential of (iii)-tree (lemma 9).

Another reason is that in the stochastic potential of (iv)-tree and (v)-tree, both ω2 and aNa+b2 are move mutations but ω2 <a+baN2 (lemma 10).

Proposition 11.

If 𝑎𝑏𝑁2 > 𝑁1 > 𝑏𝑎𝑁2, then the stochastic stable state of model with incomplete infor-mation is typical segregated state, (1,0,(1,1),(0,0)).

proof: above

4.2. Stochastic stability when type size difference is large

However, proposition 11 is based on the size of two group is close, i.e.

a

bN2 > N1 >baN2, will the stochastic stable state still be typical segregation if N1 > a

bN2 or b

aN2 > N1?

We can deduce that if N1 > abN2 or baN2 > N1 then the specification distribu-tion will change: the existence of (Bb) will distinguish and thus (iii) no longer exist.

That means the model is just like the simplified version of above model. The only difference is that we remove the specification: recurrent classes in which two group all in the block of (Bb). Moreover, the classes of (iv) and (iv) will change a little. Here

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we replace the list:

(i)Typical segregation (1,0,(1,1),(0,0))

(ii)Untypical segregation (1,0,(1,1),(1,1)) and (1,0,(0,0),(0,0)). We denote (1,0,(1,1),(1,1)) by (ii)(a) and (1,0,(0,0),(0,0)) by (ii)(b).

(iv)Recurrent classes in which two group all in the blocks of (Aa) and (Ba), or one is in (Ca) and the other is in (Aa) and (Ba).

(v)Recurrent classes in which two group all in the blocks of (Bc) and (Cc) ,or one is in (Ac) and the other is in (Bc) and (Cc).

If we set (i) as the root:

The (i)-tree

[mbNa+b2] [mbNa+b2]

(iv) (ii)(a) (ii)(b) (v)

Total N2− 1 Total N2− 1 (i)

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lemma 12.

The stochastic potential of (i)-tree is 2 [𝑚𝑏𝑁𝑎+𝑏2]+ 𝑀𝑅𝑇, where 𝑀𝑅𝑇 is the sum of move resistance of (iv) and (v).

If rooted by the relay class of (iv):

The (iv)-tree

aNa+b2+ [mbNa+b2] [mbNa+b2]

lemma 13.

The stochastic potential of (ii), (iv), and (v) are all 𝑎+𝑏𝑎𝑁2+ 2 [𝑚𝑏𝑁𝑎+𝑏2]+ 𝑀𝑅𝑇, where 𝑀𝑅𝑇 is the sum of move resistance of (iv) and (v).

To note that lemma 12 and 13 is based on the assumption N1 > abN2 or

b

aN2 > N1. Obviously the stochastic stable state is typical segregation, no matter how small the chance of move mutation is, i.e. φ does not matter, because there is no more move mutations than 𝑀𝑅𝑇 in the (i)-tree.

The relay class

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Propostion 14.

If N1 > abN2 or b

aN2 > N1, then the stochastic stable state of model with in-complete information is typical segregated state, (1,0,(1,1),(0,0)).

proof: above.

4. Conclusion

Two-group model shows another result, far from the grouping equilibrium. Equi-librium of two-group model may be segregated or non- segregate depending on the obstruction of information and misconception of players, and it also shows that with double mutations the stochastic stable state is the typical segregated state. Moreover, it solves the drawback of grouping equilibrium: move is easily to happen if people are dissatisfying on the payoff, and two-group model also shows the process of grouping formation. We can find that in the random and inherent match game, people will try to find another group to maximize his payoff. But when infor- mation is not clear on some aspects: players cannot distinct the type of each other and those in another group, or people might be "deceived" by the investigation of his or a fair institution, then non-segregation equilibrium will happen. However in a dynamic process, we will no-tice that in most situations, typical segregation is stable because only a small strategy change, the equilibrium of non-segregation will easily be broken by strategy muta-tions.

Since we know that in the dynamic model, after periods the equilibrium will be

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segregated as a typical segregation, which means efficiency for every player. Thus, what else a government should do is to increase the speed of process to form a typical segregation. Government can increase the speed by increase the number of re-matches in a given period (such like a week, a month). Once the number of match is increased, the probability of making mutation in strategy decision in a given period is increased.

From Chapter 4 we know typical segregation is more stable than other recurrent clas-ses, since the probability of mutation is increased we can infer the model is easier to derive a typical segregation, namely efficient equilibria.

Another method to solve the problem of inefficiency is removing the obstruction of information. However, it is costly for government to investigate the type of all players, and sometimes the information is always incomplete. In this situation, meth-od of increasing the number of re-matches in a given perimeth-od is a better choice.

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Appendix A.

Assume that in the same group, the proportion of players who choose to bring tennis in T.P. type is pT and that in B.P. type is pB. The proportion of T.P. players in this group is p and the size of the group is Ni.

Then for a T.P. player, he will choose tennis if

appT+ a(1 − p)pB> ap(1 − pT) + b(1 − p)(1 − pB)

For a T.P. player, he will choose tennis if

bppT+ b(1 − p)pB> bp(1 − pT) + a(1 − p)(1 − pB)

Simplify we yield

ppT+ (1 − p)pB> b a+ b ppT+ (1 − p)pB> a

a+ b

Thus for sSr−1= (1,1) or (0,0) to mutate to (1,0), we need the proportion of players who choose to bring tennis in T.P. type to change, that is, a+bb 1

p, or the propor-tion of players who choose to bring tennis in T.P. type to change, a+bb 1−p1 . (select the minimum) Consider the corresponding proportion, total population, and m memory periods, the minimum mutations are [mbNa+bi]. For (1,1) to (0,0) or (0,0) to (1,1), the minimum mutations are [maNa+bi]. For (1,0) to (1,1) or (0,0), the minimum mutations are [mNi∙ min (a+ba − p, p −a+bb )].

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